/*
 * @Author: scl
 * @Date: 2023-09-04 18:23:58
 * @LastEditTime: 2023-09-06 13:38:16
 * @Description: file content
 */
/*
 * @lc app=leetcode.cn id=79 lang=typescript
 *
 * [79] 单词搜索
 *
 * https://leetcode.cn/problems/word-search/description/
 *
 * algorithms
 * Medium (46.30%)
 * Likes:    1670
 * Dislikes: 0
 * Total Accepted:    449.9K
 * Total Submissions: 971.7K
 * Testcase Example:  '[["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]]\n"ABCCED"'
 *
 * 给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false
 * 。
 * 
 * 单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
 * 
 * 
 * 
 * 示例 1：
 * 
 * 
 * 输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word =
 * "ABCCED"
 * 输出：true
 * 
 * 
 * 示例 2：
 * 
 * 
 * 输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word =
 * "SEE"
 * 输出：true
 * 
 * 
 * 示例 3：
 * 
 * 
 * 输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word =
 * "ABCB"
 * 输出：false
 * 
 * 
 * 
 * 
 * 提示：
 * 
 * 
 * m == board.length
 * n = board[i].length
 * 1 
 * 1 
 * board 和 word 仅由大小写英文字母组成
 * 
 * 
 * 
 * 
 * 进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？
 * 
 */

// @lc code=start
function exist(board: string[][], word: string): boolean {
    const len = word.length, h = board.length, w = board[0].length, direction = [[0, 1], [0, -1], [-1, 0], [1, 0]], visited = Array.from(new Array(h), x => new Array(w).fill(true));
    const backtrack = (i:number, j:number, index:number) => {
        if (index == len)
            return true;
        visited[i][j] = false;
        let flag = false;
        for (const [x, y] of direction) {
            let newi = i + x, newj = j + y;
            if (newi >= 0 && newi < h && newj >= 0 && newj < w)
                if (board[newi][newj] == word[index] && visited[newi][newj])
                    flag = backtrack(newi, newj, index + 1);
            if (flag)
                return true;
        }
        visited[i][j] = true;
        return false;
    };
    for (let i = 0; i < h; i++) {
        for (let j = 0; j < w; j++) {
            if (board[i][j] == word[0])
                if (backtrack(i, j, 1))
                    return true;
        }
    }
    return false;
};
 // @lc code=end

